How do you integrate #int (1-x^2)/((x-9)(x+6)(x-4)) # using partial fractions?
1 Answer
#int (1-x^2)/((x-9)(x+6)(x-4)) dx =#
#=-16/15 ln abs(x-9) - 7/30 ln abs(x+6) + 3/10 ln abs(x-4) + C#
Explanation:
#(1-x^2)/((x-9)(x+6)(x-4)) = A/(x-9)+B/(x+6)+C/(x-4)#
We can determine
#A = (1-color(blue)(9)^2)/((color(blue)(9)+6)(color(blue)(9)-4)) = (-80)/((15)(5)) = -16/15#
#B = (1-color(blue)((-6))^2)/((color(blue)((-6))-9)(color(blue)((-6))-4)) = (-35)/((-15)(-10)) = -7/30#
#C = (1-color(blue)(4)^2)/((color(blue)(4)-9)(color(blue)(4)+6)) = (-15)/((-5)(10)) = 3/10#
So:
#int (1-x^2)/((x-9)(x+6)(x-4)) dx =#
#int -16/15(1/(x-9)) - 7/30(1/(x+6))+3/10(1/(x-4)) dx#
#=-16/15 ln abs(x-9) - 7/30 ln abs(x+6) + 3/10 ln abs(x-4) + C#
where